Hi Quantitative Fiance Stack Exchange,

It's my first go at GARCH models so please give me a chance with my phrasing.

I understand that GARCH models are used to forecast volatility. The GARCH(1,1) takes the form:

$$\sigma^2_t=\alpha+\beta_1\epsilon_{t-1}+\beta_2\sigma^2_{t-1}$$

I understand the lagged term $\sigma^2_{t-1}$ makes up the AR part of GARCH. However, I also understand the error term $\epsilon_{t-1}$ is dependent on the forecasting model. Consider, forecasting returns using one of the two models:

$$\hat{y_t}=\gamma\cdot y_{t-1}+\epsilon_t$$

and

$$\hat{y_t}=\theta\cdot x_{t-1}+\epsilon_t$$

Each model gives a different error term, which I believe is calculated as $\epsilon_t=y_t-\hat{y_t}$. So for the above models, error terms are $\epsilon_t=y_t-\gamma\cdot y_{t-1}$ and $\epsilon_t=y_t-\theta\cdot x_{t-1}$

Hence, is my understanding correct that calculating $\beta_1$ and $\beta_2$ of the GARCH(1,1) model depends on which forecasting model we're using?

Thank you for the help, Donny

I also understand the error term $\varepsilon_{t-1}$ is dependent on the forecasting model.

Yes, it is. The error term $\varepsilon_t$ in the GARCH model is coming from the full distributional model of $y_t$. The full model is $$ \begin{aligned} y_t &= \mu_t + \varepsilon_t, \\ \varepsilon_t &= \sigma_t \xi_t, \\ \sigma_t^2 &= \omega + \alpha_1 \varepsilon_{t-1}^2 + \beta_1 \sigma_{t-1}^2, \\ \xi_t &\sim i.i.d(0,1), \end{aligned} $$ where $\mu_t$ is the conditional mean of $y_t$, $\sigma_t^2$ is the conditional variance of $y_t$ and $d$ is some probability distribution with zero mean and unit variance.

If you are not sure which conditional mean model is best for $y_t$, you may end up with a few alternative models characterized by the conditional means $\mu_{1,t}, \mu_{2,t}, \dots$. The the corresponding error terms will differ across the models and will be $\varepsilon_{1,t} = y_t-\mu_{1,t}, \varepsilon_{2,t} = y_t-\mu_{2,t}, \dots$. This will affect the parameter estimates of the conditional variance model, just as you said.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.